$ (sec\theta - tan\theta )^2 ( 1 + sin\theta)^2 ÷ cos^2 \theta $ = ?
Answer & explanation
Correct answer: option 2
Put θ = 30º
Now , (secθ - tanθ)2 (1 + sinθ)2 ÷ cos2θ
= (√2 - 1)2 (1 + \(\frac{1}{√2}\) )2 ÷ 2
= (√2 - 1)2 (√2 +1)2
= ( 2 + 1 - 2√2) ( 2 + 1 + 2√2)
= (3- 2√2) ( 3 + 2√2)
= 9 - 8
= 1
θ